Question

A cow is tethered at point A by a rope. Neither the rope nor the cow is allowed to enter ∆ ABC.
∠ BAC = 30°
| (AB) = | (AC) = 10 m

What is the area that can be grazed by the cow if the length of the rope is 8 m?

• A. \( 134 \pi \, \frac{1}{3} \, \text{sq. m} \)
• B. \( 121 \pi \, \text{sq. m} \)
• C. \( 132 \pi \, \text{sq. m} \)
• D. \( \frac{176 \pi}{3} \, \text{sq. m} \)

Hint:

The area of a sector is calculated using the formula \( \frac{\theta}{360^\circ} \times \pi r^2 \), and make sure to adjust the radius based on the given rope length.

Answer 1

The Correct Option is D

The cow is tethered at point \( A \), and the rope forms an arc of a circle with radius 8 m, but it cannot enter triangle \( \Delta ABC \). Therefore, the grazed area is part of a sector of a circle, but we need to subtract the area of triangle \( ABC \) from it. The angle \( \angle BAC = 30^\circ \) and the radius of the sector is 8 m. The area of a sector is given by: \[ \text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2 \] where \( \theta \) is the central angle and \( r \) is the radius. For our case, we have: \[ \text{Area of sector} = \frac{30^\circ}{360^\circ} \times \pi \times 8^2 = \frac{1}{12} \times \pi \times 64 = \frac{64 \pi}{12} = \frac{16 \pi}{3} \, \text{sq. m}. \] Thus, the grazed area is \( \frac{16 \pi}{3} \) square meters.